The h Bar

5:42 AM

@schn yes there is. There is always a drop of about 0.6V across the forward biased base-emitter junction, and there is a voltage drop across the collector-base junction that depends on the base current. When the transistor is fully on the collector-base voltage drop is small.

John Rennie

6:03 AM

@schn in the diagram you mention the transistors are fully on and the voltage drop across the transistors is small compared to the voltage drop across the resistor.

Assuming you mean this diagram.

antimony

7:42 AM

cool, interesting @ACuriousMind

More Anonymous

@antimony What did your question mean? "in a high enough vacuum is it reasonable to return from statistical mechanics to kinematics?"

antimony

@MoreAnonymous well, you know how with discussion of gas laws etc. the usual analysis seems to deal with mean values because we assume there's so many particles

we can't compute their kinematic behaviour individually

More Anonymous

@antimony If you arent in the thermodynamic limit you can expect bad things to happen

antimony

oh what does that mean sorry?

More Anonymous

if u still try to apply usual thermodyanmics

antimony

7:51 AM

in high vacuum?

More Anonymous

For example the zeroth law may not hold

antimony

well i should've also acknowledged vacuum/temperature relationship, but i assume you know what i mean

ohh

when does zeroth law not hold?

More Anonymous

@antimony physics.stackexchange.com/questions/558872/…

The zeroth law only works in the thermodynamic limit as far as I understand

"In this case $H_{int} $ scales with the surface area where the systems were brought into contact, whilst $H_{QMS1} $ and $H_{QMS2} $ scale with volume, so $H_{int} $ can be neglected and your final Hamilton is the same as your initial Hamilton, in which case clearly the zeroth law holds."

antimony

ahh cool thanks!

More Anonymous

@antimony Welcome :)

More Anonymous

8:28 AM

@NiharKarve hey!

Nihar Karve

@MoreAnonymous hi

More Anonymous

I was recently wondering about this:

0

Q: Thought experiment in relativistic quantum mechanics?

Question Consider the following thought experiment in the setting of relativistic quantum mechanics (not QFT). I have a particle in superposition of the position basis: $$ H | \psi \rangle = E | \psi \rangle$$ Now I suddenly turn on an interaction potential $H_{int}$ localized at $x_0$ at time $t...

Any idea?

Nihar Karve

Interesting, lemme take a look

More Anonymous

8:42 AM

@NiharKarve I might be wrong but after thinking about it from a QFT perspective I think micro causality will come into play and demand the sudden approximation should not hold?

More Anonymous

8:53 AM

But then again $k$ can be anything so the energy scale seems to be within QM region

Nah micro causality cant be the answer

Or maybe it is?

Sigh ...

bolbteppa

Do you still disagree with (85) and (86)

It's really super simple, if you have stationary states obeying a Hamiltonian at time $t_1$, and you switch on a new Hamiltonian, the question is when does the new one start to have an effect. Obviously in principle it has an immediate effect, but still one wants to try to just ignore it and start from the state that existed at $t_1$ under the first Hamiltonian, so you look for a way to try make it work

Amazingly one can argue using time-energy uncertainty (as they do in the next few equations) that there is a small time interval in which the new Hamiltonian wont matter that much so the new effects will be very minor, this has absolutely nothing to do with how many particles are in the system or much about the Hamiltonian beyond using stationary states

More Anonymous

@bolbteppa I was disagreeing about (86) to (87)

And are talking about this
( https://chat.stackexchange.com/transcript/message/56764808#56764808 )

bolbteppa

But there's nothing wrong with (86) and (87), your argument the other day is irrelevant

More Anonymous

In QM or QFT is it irrelevant?

bolbteppa

You're not even explaining why it doesn't apply to QFT

More Anonymous

9:07 AM

I think your right ... It should work

Im not familiar enough with QFT to have the same confidence as you

@bolbteppa any idea about the other question?

though

bolbteppa

If you see in (84) they basically model the change from one Hamiltonian $H_1$ for $t < t_1$ to a new Hamiltonian $H_2$ for $t > t_2$ as an intermediate $H(t)$ for $t_1 < t < t_2$, and examine cases where the change is fast (sudden approximation) and slow (adiabatic approximation) so in the fast case, if the intermediate Hamiltonian only acts for a time bound by time-energy,

the effect on the state under $H_1$ by $H(t)$ will be negligible and then since a new $H_2$ kicks in and lasts for longer this will start having a dominant effect so you basically have $|\Psi(t_2)> = |\Psi(t_1)>|_{t_2} + \text{error}$ if that makes sense

More Anonymous

It does work out if one state can be written as a linear combination of another (which was my objection) ... I wasn't sure if this was the case in QFT with changing particle numbers

bolbteppa

If you write wave functions in an occupation number basis, it doesn't matter if particle numbers start changing the coefficients account for that because you sum over all possible states so the only question is whether a given energy state is occupied and if so by how many particles

This is why qft is done via second quantization

More Anonymous

9:24 AM

Yes this was what I was uncertain about. Thanks for clarifying :)

bolbteppa

I have no idea about the other question I'd say you'd have to formulate it in these terms

Schrödinger functional

In mathematical physics, some approaches to quantum field theory are more popular than others. For historical reasons, the Schrödinger representation is less favoured than Fock space methods. In the early days of quantum field theory, maintaining symmetries such as Lorentz invariance, displaying them manifestly, and proving renormalisation were of paramount importance. The Schrödinger representation is not manifestly Lorentz invariant and its renormalisability was only shown as recently as the 1980s by Kurt Symanzik (1981). Within the Schrödinger representation, the Schrödinger wavefunctional...

More Anonymous

@bolbteppa I think I could possibly formulate it in terms of normal QFT as well

Or maybe not

FakeMod

Directional derivative of a function $f$ along a curve $\Gamma$ at a point $\mathbf x$ do not necessarily equal to the directional derivative of $f$ along the tangent vector of $\Gamma$ at the point $\mathbf x$. This statement is correct, right?

Slereah

9:40 AM

Doesn't it?

FakeMod

@Slereah I mean, I have a counter example of a non-well behaved function. $$f(x,y) = \begin{cases} 0 & x = y = 0 \text{ or } y \neq x^2 \\ 1 & \text{otherwise}\end{cases}$$

Slereah

That doesn't seem continuous

FakeMod

@Slereah Yeah, but directional derivatives do exist along all the directions towards the origin.

But the directional derivative does not exist along the curve $y=x^2$.

Despite existing along $y=x^2$ curve's tangent vector at origin.

But again, I am not exactly sure so I came here to ask. I feel that if there can be a case where the directional derivative exists when approached through the tangent and does not exist when approached through the curve itself, then there possibly could be an example of a (possibly continuous) function where both the values exist and differ.

Multivariable calculus has removed all prejudices and assumptions from my mind. Now my mind thinks anything is possible in multivariable calculus.

Slereah

10:00 AM

Calculus isn't Christ

6

bolbteppa

When you say approached through the tangent you mean approached through a curve given as a line

FakeMod

@bolbteppa Yes.

bolbteppa

Directional derivatives don't point along curved directions like parabolas

FakeMod

@bolbteppa Yeah, but can't we define directional derivatives along a curve, though they are not very commonly found. In fact, I found quite a few questions on MathSE asking to compute the directional derivatives of functions along curves.

bolbteppa

Link

Slereah

10:12 AM

A curve defines a direction

by its tangent

FakeMod

0

Q: Directional Derivative along A Curve

Find the directional derivative of $f(x,y,z) = x^2yz^3$ along the curve with parametric equations $$\begin{align}x & = e^{-t}, \\ y & = 1 + 2 \sin t, \\ z & = t - \cos t, \end{align}$$ at the point $P$ where $t = 0$. I know how to find the directional derivative along a vector, but how I can I f...

real-analysis multivariable-calculus

Slereah

If you compute the derivative of the pullback of a function along a curve, that is the directional derivative

FakeMod

@Slereah But according to that logic, a limit at a point would exist if it exists when the point is approached from all directions (tangents), because after all every other curve (according to this statement) can be viewed as a tangent. But we know that the limit may not exist in such a case.

Slereah

I mean, the directional derivative may not exist in all directions

yes

If you want to define all derivatives at a point, you have to deal with equivalence classes of every curve at that point

that's one way to define the tangent space

bolbteppa

The answer finds the tangent to the curve i.e. the direction in which one computes the directional derivative, so in your first statement above it doesn't make sense to draw a distinction

Slereah

10:17 AM

You may remember for instance that not all complex functions are holomorphic, ie have a derivative in every direction

famously the complex conjugate isn't holomorphic

FakeMod

@bolbteppa That answer was the exact reason why I questioned what that answer took for granted.

@Slereah Yes, I do remember that.

@Slereah I am looking up what equivalence classes are. I will reply once I understand what they are. Thank you for your patience.

bolbteppa

One still always has to find the tangent vector pointing in some direction to compute the directional derivative, just because you want to move along some curve like a parabola in the plane you still need to begin by moving in some linear direction to do that which is all the derivative cares about

If a function doesn't have a directional derivative in some direction big whoop, that's too bad for the function, it's like analyzing a single variable function which has no derivative. The non-trivial thing in these higher dimensions is that one could have directional derivatives in all directions but the notion of 'differentiability' as a concept intrinsic to the dimension you're at can still fail

More Anonymous

@all is this question wrong? I find it concerning when I receive a downvote under 30 views :(( physics.stackexchange.com/questions/608003/…

ACuriousMind

10:32 AM

@MoreAnonymous The "please don't post your questions here shortly after they're posted" rule goes for you, too.

FakeMod

@bolbteppa But can't we parametrize the curve $\Gamma$ (let's say we use $t$ as our parameter, thus the curve becomes $\Gamma (t)$) along which we're gonna take the derivative of our function $f$ and the find the derivative as follows $$f'(\mathbf x (t_0)) = \lim_{h\to 0}\frac{f(\mathbf{x}(t_0 + h)) - f(\mathbf x (t_0))}{h}$$ Now I don't see any apparent reason for this derivative to equal the directional derivative corresponding to the tangent along $\Gamma (t)$ at $t_0$.

More Anonymous

@ACuriousMind Wasnt aware of that rule

My bad

Is there any room where I can ask for constructive feedback?

I believe MSE has one

ACuriousMind

If everyone who gets a downvote asked about it here the room would consist of little else

and if the downvoter had wanted to explain themselves to you they'd have left a comment in the first place

More Anonymous

@ACuriousMind which is why it would be wonderful to create a room specifically for that purpose

ACuriousMind

@MoreAnonymous ...creating the room does not automatically spawn people willing to spend their time there

bolbteppa

10:36 AM

@FakeMod you can set $g(h) = f(\mathbf{x}(t_0 + h))$ and Taylor expand the single-variable function in $h$, so the numerator is $g(h) - g(0) = g(0) + g'(0)h + ... - g(0) = g'(0) h + \mathcal{O}(h^2)$, the $g'(0)$ is the directional derivative

ACuriousMind

I understand that the people being downvoted would flock to such a room to ask about their questions, but where do you get the people willing to give feedback from?

More Anonymous

@ACuriousMind Somehow the goodwill of the people prevailed in MSE

ACuriousMind

I mean, you can just create a room and see if people show up, but chances are, they won't

More Anonymous

How many points do I need?

ACuriousMind

To create a chat room? Only 100 rep.

More Anonymous

10:38 AM

I for one wouldnt mind helping people out (if it is a topic within my ablity)

FakeMod

@bolbteppa Hmm... This makes the picture clearer. According to this, I can conclude that there will never be a differentiable multivariable function which will have different directional derivatives when taken along the curve and a tangent. But what in the cases of multivariable functions which are continuous but aren't differentiable at the point under consideration.

As for the functions which are not continuous at the point under consideration, I already gave a counter example, though in that example, the directional existed in only one of the cases.

But we can easily find a non-continuous function in which directional derivatives exist in both the cases and still differ.

So the only type of functions which are unresolved seem to be the continuous but non-differentiable class of functions. What would hold true in their case?

More Anonymous

@ACuriousMind Thanks and magic poof: chat.stackexchange.com/rooms/118530/constructive-feedback

And they say things don't pop outta nowhere :P

ACuriousMind

@FakeMod I think your idea that there are distinct concepts of a directional derivative "along a curve" or "along the tangent to a curve" is wrong. There is just one concept - you have a differentiable function and a curve and you apply the tangent vector field to the curve to the restriction of the function to the curve. This is the same as pulling back the function along the curve $\gamma : \mathbb{R}\to M$ and then computing its only derivative there

$\gamma$ is necessarily an embedding in this definition. I suspect that the curves you are trying to use in the cases where you get some sort of derivative for non-differentiable functions is not

ACuriousMind

11:06 AM

For instance, I don't really understand your example above. Your function is $1$ along the curve $y = x^2$ except at $x=0$, and zero everywhere else. Its pullback along $t\mapsto (t,t^2)$ is the function $\bar{f}(t)$ that is zero at $t=0$ and 1 everywhere else. It is not differentiable at $t=0$, because it is not continuous.

What seems to be confusing you is that you think the "derivative along the curve" at zero should be the same as the derivative of $f$ along the vector field $\partial_x$ in a neighbourhood of zero. This is just not the case here, and the post you're quoting is really just missing the stipulation that the function must be differentiable for this to hold.

but no one really thinks of "derivative along the tangent to a curve" as a distinct concept, it's just how one computes the derivative along a curve for differentiable functions

FakeMod

@ACuriousMind True. For well behaved functions, I completely agree that both the concepts are the same, but in the case where the function is not differentiable, but it is continuous, would we still be able to call both the definitions equivalent?

ACuriousMind

Depending on how you define "tangent vector", you get in trouble for directional derivatives for non-differentiable functions anyway

if you define a tangent vector as an equivalence class of curves, then the definition of what it means to apply the tangent vector as a derivative to a function is to do the "along the curve" computation above, which doesn't work for a non-differential function

i.e. whether or not your notion of the two computations being different even makes sense depends on how you define "tangent vector" to begin with

since differential geometry wants all definitions of tangent vector to be the same (lots of other things break otherwise), you probably should view the directional derivative as being ill-defined for functions that are not totally differentiable

FakeMod

@ACuriousMind I see :) But this has raised another question. If a directional derivative of a function exists at a point when taken along all possible differentiable curves, would the function necessarily be differentiable at that point?

@ACuriousMind Oh, I get what you mean.

@ACuriousMind Well then the "another question" of mine already been answered.

Thank you @Slereah @bolbteppa @ACuriousMind See you later.

schn

11:25 AM

@JohnRennie thanks for the reply. Yes, that’s the diagram I meant. Since, suppose we connected a LED at the top before the two transistors and say A was on (LED should not light up due to the AND truth table). However if there is a voltage drop over collector and emitter, current would flow and light up the LED, right?

@JohnRennie Of course, OUT is at the bottom in the figure, but could also be removed to the top along with resistor, right?

@JohnRennie As shown here around 5:22 youtu.be/sTu3LwpF6XI

John Rennie

@schn If you put the resistor at the top you get a NAND gate i.e. the output is high when either or both of the inputs are low, and the output is low when both inputs are high.

schn

@JohnRennie Is then the schematic shown in the video wrong for the AND gate the author has constructed?

@JohnRennie There the resistor is at the top.

John Rennie

@schn I think he means the AND gate is true when the LED is on. He isn't using an output voltage, just whether or not the LED is illuminated.

If we took an output voltage from the bottom of the resistor then he would indeed have a NAND gate not an AND gate.

schn

11:40 AM

Yes. So in my original figure and the one you showed here, the LED represents the resistor, correct?

John Rennie

No, the LED has only a very small resistance. It just shows when a current is flowing. In the video the circuit has a resistor as well as the LED, and that resistor does the same as the resistor in this circuit.

schn

@JohnRennie I see. Of course...It is a bit confusing, since his schematic looks different. If the voltage drop over the collector and emitter of a transistor is assumed to be negligible, then his schematic makes sense, since then there is only a voltage drop over the LED and resistor (and consequently current flow) if both A and B are on.

John Rennie

@schn Yes

schn

@JohnRennie So it assumed to be negligible, the voltage drop over collector and emitter? :) There is of course the voltage drop of 0.6 over base and emitter in order for the transistor to "turn on".

John Rennie

Yes

schn

11:53 AM

@JohnRennie Although, upon closer thought, in his schematic (will upload picture), when A is on, current can flow throught transistor A (collector to emitter), isn't there then a voltage drop over the resistor?

John Rennie

If both A and B are high then both transistors conduct, so the voltage drop across both transistors is small. That means most of the voltage will be dropped across the resistor i.e. where will be about 4V across the transistor and about 1V across the two transistors.

@schn OK so far?

schn

@JohnRennie Sure.

John Rennie

So there will be a current I = 4/R flowing in the circuit. That current will cause the LED to light up.

schn

@JohnRennie But suppose only A is high, then most of the voltage drop will also be over the resistor, right? Shouldn't that make the LED light up as well, since there is a current flow?

John Rennie

No. If only A is high transistor B will be off i.e. it has an effectively infinite resistance. That means the whole 5V will be dropped across transistor B so the voltage dropped across the resistor is zero.

schn

12:01 PM

@JohnRennie Clear!

John Rennie

:-)

More Anonymous

12:50 PM

@ACuriousMind I've edited the question. The potential mentioned is no longer unphysical

And if it's okay with you can you delete that comment since (if you agree) it is no longer relevant. People get the impression that there question is flawed in a way it is not

More Anonymous

1:04 PM

?

RewCie

1:32 PM

@schn This is and gate..

If both A and B are activated, then only the current will dare to flow.

ACuriousMind

2:40 PM

@MoreAnonymous Please flag such comments as "no longer needed", then one of the mods will clean them up. Chat isn't the right venue for this.

More Anonymous

3:00 PM

@ACuriousMind Wasn't aware of this procedure and thought flagging was only for abusive comments or something.. Will do from on :)

polite proofs

3:15 PM

@Semiclassical Have you read any of Kleppner's books?

For mechanics?

Semiclassical

sounds familiar but i've not seen them myself

but maybe someone else here has

polite proofs

What about David Halliday's Volume I or Volume II?

Semiclassical

nope

polite proofs

One moment

amazon.com/Physics-Halliday-Kenneth-Robert-Resnick-ebook/dp/…

Can you have a look at the ToC and tell me what you think?

Semiclassical

same Halliday as Halliday and Resnick?

polite proofs

3:16 PM

Yes

Semiclassical

neat

polite proofs

Right

Semiclassical

seems like the standard kinematics -> forces -> energy sequence

polite proofs

Is it a quality book? I am looking for a quality physics textbook that isn't afraid of math and doesn't have an awful lot of useless illustrations just to get the page number up

Semiclassical

could not tell you based on the ToC

i don't know a lot of the intro physics books. perhaps others here do

polite proofs

3:27 PM

(Others feel free to let me know what you think)

bolbteppa

4:13 PM

Kleppner is more math heavy than the other one

polite proofs

@bolbteppa Really? That's surprising. Have you read either?

bolbteppa

Yes

See the amazon reviews for them for example

polite proofs

@bolbteppa Which one have you read?

bolbteppa

Big parts of both (well the other Halliday, they're all pretty similar), try this for some advice

polite proofs

@bolbteppa Title does not apply to me. Funnily enough, I know some real analysis (a few chapters into Baby Rudin), but I don't know any physics.

By the way

bolbteppa

4:25 PM

You should probably get

Fundamentals of Physics

Fundamentals of Physics is a calculus-based physics textbook by David Halliday, Robert Resnick, and Jearl Walker. The textbook is currently in its eleventh edition (published 2018). The current version is a revised version of the original 1960 textbook Physics for Students of Science and Engineering by Halliday and Resnick, which was published in two parts (Part I containing Chapters 1-25 and covering mechanics and thermodynamics; Part II containing Chapters 26-48 and covering electromagnetism, optics, and introducing quantum physics). A 1966 revision of the first edition of Part I changed the...

polite proofs

Don't mistake Resneck's Physics book with Volume I of Physics

Volume I of Physics is used by the USA IPhO team at the training camp

They're entirely different books

Why that over Volume I?

bolbteppa

They use calculus and the problems in there come up in the GRE sometimes almost word-for-word

polite proofs

Volume I is also Calculus based

@bolbteppa I am not trying to challenge what you said, by the way, I am just really curious why you would recommend that over Volume I

Since committing to a thousand page textbook is not something I should take very lightly, imo :-)

bolbteppa

Yeah I think the Physics one organizes things a bit better (e.g. putting energy later) and it's probably better than the Fundamentals one overall but I'd say the difference is not that big, though I am not 100% sure, there's a few of discussions on physicsforums on the difference between them

polite proofs

I don't mind harder math, fyi

I actually would prefer that

Since that usually means the authors didn't have to "hold back" as much

danielunderwood

4:42 PM

If you don't mind math, there's always the intermediate level books (Taylor/Marion, Griffiths)

polite proofs

@danielunderwood The issue is, they usually assume knowledge of physics

danielunderwood

Though I was too lazy to look at my intro-level textbooks, so I don't know what additional details they may offer

polite proofs

Not harder math

bolbteppa

Maybe get the Physics one then if it's used by IPhO, if you find it it easy you can make it harder by doing IPhO problems, the fact it uses calculus means it's not going to be trivial, more or less everything is derived from first principles if you can finish it then you'd be able to read advanced ones which to some extent will be re-writing some of this stuff more generally more concisely assuming you know the basics as a book like this gives

polite proofs

Okay. One more thing: Why does something like 8.012 (Honors Mechanics at MIT) focus so much on.... mechanics? That is, in general they have a big focus on mechanics

While something like Volume I goes over a lot of things

And mechanics is only a few chapters

bolbteppa

4:54 PM

In the early chapters it's because one considers harder problems and a more general mathematical perspective, I don't think a book like Halliday even emphasizes things like constraints the way a book like Kleppner does

polite proofs

What kind of constraints?

bolbteppa

How to write constraint equation for a pulley system - 1 [IITJEE]

►

polite proofs

Oh I see

polite proofs

5:15 PM

i.imgur.com/HxrwxB3.png Why does this guy use modular arithmetic equivalence/logical equivalence for this?

Oh no... Physicists, why?

ACuriousMind

@politeproofs Some people use $\equiv$ to mean definition, it's not specific to physics

polite proofs

@ACuriousMind I have never seen $\equiv$ used to mean $\stackrel{\text{def}}=$

Can't think of a single book

ACuriousMind

Notation varies, now you've seen it :P

polite proofs

(In a math textbook, that is)

JMac

5:41 PM

@politeproofs I don't remember seeing $\stackrel{\text{def}}=$ and have seen $\equiv$ used quite a bit

ACuriousMind

$:=$ is also popular

polite proofs

:= is fine

@JMac What, in a math book?

JMac

@politeproofs Probably, yeah. I've definitely seen it in physics contexts like your picture was.

polite proofs

@JMac Physics contexts don't matter.

JMac

You linked to a picture that was in a physics context, so I'd say using notation common to physics context makes a lot of sense...

polite proofs

5:54 PM

@JMac I claim that physicists should not butcher notation like that. ;)

Semiclassical

different does not mean wrong

JMac

@politeproofs The wikipedia article on the triple bar symbol gives an example where they are the same. "This symbol is also sometimes used in place of an equal sign for equations that define the symbol on the left-hand side of the equation, to contrast them with equations in which the terms on both sides of the equation were already defined. An alternative notation for this usage is to typeset the letters "def" above an ordinary equality sign, ...

"$a \stackrel{\text{def}}= b$"

I don't think different symbols used are a big deal as long as you are consistent and clearly define them before you start using them.

polite proofs

How do I disable the RSS feed taking up so much of my screen every so often as a popup?

Semiclassical

6:10 PM

@JMac the physics notation which I know drives math people crazy is $f(x,y)=x^2+y^2\implies f(r,\theta)=r^2$

ZeroTheHero

7:34 PM

yet, quoting D'Alembert: "Allez en avant, et la foi vous viendra."
(Forward! and faith will come.) [advice to those who questioned the calculus] Quoted in A L Mackay, Dictionary of Scientific Quotations (London 1994)

fqq

9:09 PM

@Semiclassical I'm a physicist and it drives me crazy too

I do use it sometimes though

Semiclassical

:D

fqq

you could do worse, like using the same name for a function $f(x)$ and its Fourier transform $f(q)$.

Semiclassical

ew

yeah, that's a bridge too far for me

at least the $f$ case has the excuse of "i'm measuring a given quantity at a particular point regardless of how i label the point"

so calling the electric potential $\Phi$ regardless of whether you express it in Cartesian or spherical coordinates

fqq

9:30 PM

you could argue along the lines that the Fourier transform is the "same" function in a different basis

(not that I want to do that)

Andrew Micallef

10:22 PM

hi there, would anyone have a moment to talk me through the maths in [this](https://physics.stackexchange.com/a/418456) answer?
https://physics.stackexchange.com/a/418456

I'm unclear about what happens to the dimensions of the box and am unfamiliar with the terminology : $$\mathrm{x=0 \ face:} \ \mathbf K_1$$ etc

I want to make some python code for modelling a bar magnet and thought this would be a nice place to start.

I'm unclear about what happens to the dimensions of the box and am unfamiliar with the terminology : $\mathrm{x=0 \ face:} \ \mathbf K_1$ etc

I want to make some python code for modelling a bar magnet and thought this would be a nice place to start.

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